Problem: Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} y & x - 1 \\ \\ y - 1 & x \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(2, -1)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely
Explanation: The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} y & x - 1 \\ \\ y - 1 & x \end{bmatrix} \right) \\ \\ &= xy - (x - 1)(y - 1) \\ \\ &= xy - (xy - x - y + 1) \\ \\ &= x + y - 1 \end{aligned}$ If we evaluate $|J(f)|$ at $(2, -1)$, we get $0$. Because the Jacobian determinant here is equal to $0$, we can conclude that $f$ will infinitely contract the space around $(2, -1)$. To recap, the Jacobian determinant of $f$ is $x + y - 1$, and $f$ will infinitely contract the space around the point $(2, -1)$.